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Why work is defined as a scalar?

  1. Jun 13, 2015 #1
    we know work is F.s which is a scalar quantity. but why it is defined this way??? work actually has something to do with direction. Negative work means reduction in velocity and velocity has direction. What was the problem of defining work as vector that lies in the direction of displacement??
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  3. Jun 13, 2015 #2


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    It does not.
    It does not have to be a reduction in velocity, it can be everything. Also, you can change the direction of velocity without altering speed. Do you want to have a separate work quantity for every possible direction where deceleration can occur?
    It does not make sense because work does not care about the direction of motion of whatever object was used. It's like assigning a temperature a material property based on the material used to measure it. "It is 30 degrees mercury-celsius today" - "oh, according to my thermometer it is 30 degrees semiconductor-celsius".
  4. Jun 13, 2015 #3


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    The quantity F.s is a useful quantity. Because it is useful, it is used often, and it was therefore convenient to give it a name. There really isn't any reason for a definition other than usefulness and convenience.

    You mean something like ##|\mathbf{F}|\mathbf{s}##? Where is such a quantity used? I cannot think of a single time I have ever seen that quantity used.
  5. Jun 13, 2015 #4
  6. Jun 15, 2015 #5
    Let's look at work as force acting over a distance.

    First the force vector:

    Fortunately it has units of work, not force, and everything makes sense.

    [itex]W[ newton \cdot meter ] = F_x[newton] dx[meter] + F_y[newton] dy[meter] + F_z[newton] dz [meter][/itex]

    Normally, you are taught that [itex]F = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} [/itex].

    i, j and k are unit vectors in the x, y, and z directions, respectively. The little dx, dy, and dz replace these.
  7. Jun 15, 2015 #6
    Force is a "covector" having units of work. The vector elements have units of force.

    [tex]w = F_x dx + F_y dy + F_z dz[/tex]
    Integrated over a spatial interval, [itex]W_{ab} = \int_{a}^{b} w [/itex]
  8. Jun 15, 2015 #7


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    Work is supposed to be a measure of how "hard" it is to do something. If work had been a vector in the direction of the displacement, then if you push a heavy object 5 meters in one direction and then push it back, you will have performed zero work.

    Also, what if you push an object in a curved line? What would be the direction of the work vector then?
  9. Jun 15, 2015 #8


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    Mathematically, work is the dot product between the force and distance. Dot product creates a scalar out of two vectors. Note that if you push the side of a truck that happens to be moving forward (perpendicular to your force), then you aren't doing work on the truck. The dot product converts the two directions (your force and the trucks motion) into a single quantity, independent of direction, since you can rotate the entire coordinate system and you still do no work on the truck.

    Work shares the same units as energy, and energy has no direction. In fact, in special relativity, energy and momentum form a 4-vector. You could think of the energy as the component of energy-momentum directed in the time direction, while the momentum is the three components directed in space directions.
  10. Jun 15, 2015 #9


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    What? That makes no sense. Force has units of mass*length/time^2. Work has units of mass*length^2/time^2.
    Note that length^2 = length dot length, which is a scalar.
  11. Jun 15, 2015 #10
    [itex]W[ML^2T^{-2}] = F_i[MLT^{-2}]dx^i[L][/itex]

    [itex]W_{ab} = \int_a^b F_i dx^i = \int_a^b W[/itex]

    Re. length squared as a scalar. Interesting.

    [itex]\frac{\partial}{\partial x} dx[/itex] is scalar and dimensionless.
    Last edited: Jun 15, 2015
  12. Jun 18, 2015 #11
    Work is a quantity that just tell you how much the force achieved its "Goal"(the goal here is the displacement from one place to another with the direction of the force ).
    suppose that you are making a force on a block,but the block didn't do displacement at all.Then in this case the work is zero(you didn't achieve your goal).
    If you want to multiply vectors like how you you multiply real numbers,the two vectors must be in same direction else you should project one into another(and that what scalar product does).
    W=F.r, you should project F in the direction of r because it is the condition where your goal is achieved or oppositely achieved (if your force is projected in the same direction but in opposite sense with displacement).
    F is the force and r is the displacement.If the block is displaced and there is zero force then work is zero,and if there is forced applied on the block but no displacement occurred then work is zero either.
    SO WORK CAN'T BE A VECTOR QUANTITY.IT JUST TELL YOU HOW MUCH YOUR GOAL IS ACHIEVED. Look at the figure below. 20150618_110957.jpg
    Last edited: Jun 18, 2015
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